- #1
goldfish9776
- 310
- 1
Member warned about proper use of the template and showing an attempt at solution.
In summary: Very frustrating.No, see Doc Al response above for clarification. If the block is moving, the friction force is just \mu_k(mgcos\theta) . You have to charitably reinterpret the problem. Very frustrating.In summary, the conversation discusses finding the total force acting downwards on a plane and whether or not to consider the tension in point B and how to calculate the tension on the rope. However, there is confusion about the problem statement as it implies that the block is in equilibrium but also provides a value for kinetic friction, which is contradictory. It is suggested to reinterpret the problem to find a solution.
- #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
- 12,811
- 1,673
goldfish9776 said:
Do you feel that the tension has no effect on the block?
The block is in equilibrium as shown. How does the block stay in equilibrium?
- #3
Doc Al
Mentor
- 45,588
- 2,314
It's a force acting on the block, so you'd better.goldfish9776 said:
You are told that point B is in equilibrium.goldfish9776 said:
- #4
PhanthomJay
Science Advisor
Homework Helper
Gold Member
- 7,201
- 530
tension and frictiongoldfish9776 said:
try a free body diagram of joint B. Please show your work. Is the block in equilibrium?
Homework Equations
The Attempt at a Solution
[/QUOTE]- #5
- #6
PhanthomJay
Science Advisor
Homework Helper
Gold Member
- 7,201
- 530
No. But there seems to be a problem in the problem statement, because it states that joint B is in equilibrium which implies that the block must be in equilibrium also. But the solution gives a kinetic (moving) friction value, so something went wrong here, the joint cannot be in equilibrium with the given values.goldfish9776 said:
Last edited by a moderator:
- #7
goldfish9776
- 310
- 1
so , the correct one should be 100(9.81)sin20 +150 only ?PhanthomJay said:
- #8
PhanthomJay
Science Advisor
Homework Helper
Gold Member
- 7,201
- 530
No, a free body diagram of the block will show that in addition to the two down-the-plane forces you have identified, there is a tension force acting up the plane and possibly a friction force acting parallel to the plane. Now the problem states that the joint B is in equilibrium (not accelerating) and thus the block also must not be accelerating. But when you draw a free body diagram of the joint at B to determine the tension force, and apply that force to the block, there is not enough available static friction force down the plane to keep the block from moving, which is in contradiction with the problem statement. As I see it, the problem is not correctly worded.goldfish9776 said:
- #9
Doc Al
Mentor
- 45,588
- 2,314
I agree. A very poorly worded problem. (It reads as if translated from another language.)PhanthomJay said:
You can certainly get the given answer, but you have to disregard (or charitably reinterpret) some of the problem statement. You first assume equilibrium, then ask if that is possible. It's not--as PhanthomJay explains--so the friction must be kinetic.
- #10
goldfish9776
- 310
- 1
isn't it = 100(9.81)sin20 ?PhanthomJay said:
- #11
PhanthomJay
Science Advisor
Homework Helper
Gold Member
- 7,201
- 530
No, see Doc Al response above for clarification. If the block is moving, the friction force is just [itex]\mu_k(mgcos\theta) [/itex]. You have to charitably reinterpret the problem.goldfish9776 said:
FAQ: Total forces acting downward on the plane
What are total forces acting downward on the plane?
Total forces acting downward on the plane refer to the combined forces that are pushing or pulling the plane towards the ground. These forces include the weight of the plane, air resistance, and gravity.
How are total forces acting downward on the plane calculated?
Total forces acting downward on the plane can be calculated by adding all the individual forces acting on the plane in a downward direction. This includes the weight of the plane, which is determined by its mass and the acceleration due to gravity, as well as any other external forces acting on the plane.
What factors affect the total forces acting downward on the plane?
The total forces acting downward on the plane can be affected by various factors such as the weight and mass of the plane, air density, altitude, and velocity of the plane. These factors can change the magnitude and direction of the forces acting on the plane.
Why is it important to consider the total forces acting downward on the plane?
Considering the total forces acting downward on the plane is crucial for ensuring flight safety. If these forces are not properly balanced, it can affect the stability and control of the plane, potentially leading to accidents. It also helps in determining the performance and efficiency of the plane.
How can the total forces acting downward on the plane be controlled?
The total forces acting downward on the plane can be controlled by adjusting the plane's weight, angle of descent, and velocity. Properly maintaining and monitoring the plane's systems and aerodynamics can also help in controlling these forces. Pilots also play a crucial role in managing these forces through their flying techniques.